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natka813 [3]
3 years ago
10

Until recently, hamburgers at the city sports arena cost $ 2.50 each. The food concessionaire sold an average of 1750 hamburgers

on game night. When the price was raised to $ 3.10 hamburger sales dropped off to an average of 1,450 per night. The? concessionaire's fixed costs were $1,903.00 per night and the variable cost was $1.44 per hamburger. Answer the following questions (A) through (F).(A) Assume that the relationship between price p and demand x is linear. Express p as a function of x and find the domain of this function.
p=The domain of p is(Type a compound? inequality.)
(B) Find the revenue function in terms of x and state its domain.- ?R(x)=The domain of R(x) is(Type a compound? inequality.)(C) Assume that the cost function is linear. Express the cost function in terms of x.C(x)=(D) Graph the cost function and the revenue function in the same coordinate system.Find the break-even points.The break-even points are(Simplify your answer. Type an ordered pair. Use a comma to separate answers as? needed.)(E) Find the profit function in terms of x.P(x)=(F) Evaluate the marginal profit at x=600 and interpret the results.The marginal profit at x=600 is $.b)At a production level of 600 ?hamburgers, the profit is increasing at a rate of $ per hamburger.a)At a production level of 600 ?hamburgers, the profit is decreasing at a rate of $ per hamburger.

Business
1 answer:
Artyom0805 [142]3 years ago
6 0

Answer:

  (A) p = -0.002x +6; 0 ≤ x ≤ 3000

  (B) R(x) = x(6 -0.002x); 0 ≤ x ≤ 3000

  (C) C(x) = 1.44x +1903

  (D) (550, 2695), (1730, 4394.20)

  (E) P(x) = -0.002x^2 +4.56x -1903

  (F) increasing at $2.16 per hamburger

Explanation:

(A) The two-point form of the equation for a line can be used.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

The two points we have are ...

  (x, p) = {(1750, 2.50), (1450, 3.10)}

so the equation is ...

  p = (3.10 -2.50)/(1450 -1750)/(x -1750) +2.50

  p = 0.6/-300(x -1750) +2.50

  p = -0.002x +6

The domain of this function is where x and p are greater than 0. That will be for ...

  0 ≤ x ≤ 3000

__

(B) Revenue is the product of burgers sold (x) and their price (p).

  R(x) = xp

  R(x) = x(6 -0.002x)

The domain of R(x) is 0 ≤ x ≤ 3000. This is the same as the domain of p(x).

___

(C) The cost function is the sum of fixed costs and variable costs:

  C(x) = 1.44x +1903

__

(D) See the attachment for a graph of cost and revenue. The break-even points are (x, revenue) = (550, 2695), (1730, 4394.20).

__

(E) Profit is the difference between revenue and cost.

  P(x) = R(x) - C(x) = x(6 -0.002x) -(1.44x +1903)

  P(x) = -0.002x^2 +4.56x -1903

__

(F) The marginal profit is the derivative of the profit function:

  P'(x) = -0.004x +4.56

  P'(600) = -0.004(600) +4.56 = -2.40 +4.56 = 2.16

At a production level of 600, the profit is increasing at a rate of $2.16 per hamburger.

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The receipt of dividends and interest from abroad as a result of ownership of foreign assets by a country's residents is recorde
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Pooler Corporation is working on its direct labor budget for the next two months. Each unit of output requires 0.15 direct labor
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Answer:

$13,335

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Required production in units for April and May are 6,500 units and 6,200 units respectively.

Direct labor hours needed is 0.15 for both months.

Total direct labor hours needed for each month would be;

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